Sequences and Series – Exercise 9.3[11-20] – Class XI

11. Evaluate 

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12. The sum of first three terms of a G.P. is ³⁹/₁₀ and their product is 1. Find the common ratio and the terms.

Solution:

Let ^a/_r, a, ar be the first three terms of the G.P.

^a/_r + a + ar = ³⁹/₁₀ —-(1)

(^a/_r)(a)(ar) = 1 —–(2)

From (2), we obtain a³ = 1

a = 1

Substuting a = 1 in equation (1), we get

¹/_r + 1 + r = ³⁹/₁₀

1 + r + r² = ³⁹/₁₀ r

10 + 10r + 10r² – 39r = 0

10r² – 29r + 10 = 0

10r² – 25r – 4r + 10 = 0

5r(2r – 5)-2(2r – 5)= 0

(5r – 2)(2r – 5) = 0

r = ²/₅ or ⁵/₂

Thus the three terms of G.P. are ⁵/₂, 1 and ²/₅

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13. How many terms of G.P. 3, 3²,3³ … are needed to give the sum 120?

Solution:

The given G.P. is 3, 3², 3³,…

Let n terms of this G.P. be required to obtain the sum as 120.

Thus, four terms of the given G.P. are required to obtain the sum as 120.

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14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum of n terms of the G.P.

Solution:

Let the G.P. be a, ar, ar²,…

According to the given condition,

a + ar + ar² = 16 and ar³ + ar⁴ + ar⁵ = 128

⇒ a (1 + r + r²) = 16 ……………. (1)

ar³(1 + r + r²) = 128 …………….. (2)

Dividing equation (2) by (1), we obtain

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15. Given a G.P. with a = 729 and 7^th term 64, determine S₇

Solution:

a = 729 and a₇₌ 64

Let r be the common ratio of the G.P. It is known that

a_n = a.r^n-1

a₇ = a.r^7-1 =(729)r⁶

64 = 729(r⁶)

r⁶ = ⁶⁴/₇₂₉

r⁶ =(²/₃)⁶

r = ²/₃

Also it is known that,

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16. Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

Thus the required G.P. is –⁴/₃, –⁸/₃, –¹⁶/₃,… or 4, -8, 16, -32,…

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17. If the 4^th ,10^th and 16^th terms of a G.P. are x , y and z respectively. Prove that x , y, z are in G.P.

Solution:

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a₄ = ar³ = x ———-(1)

a₁₀ = ar⁹ = y ———-(2)

a₁₆ = ar¹⁵ = z ———-(3)

Dividing (2) by (1), we get

^y/_x = (ar⁹)/(ar³)

^y/_x = r⁶

Dividing (3) by (2), we get

^z/_y = (ar¹⁵)/(ar⁹)

^z/_y = r⁶

^y/_x = ^z/_y

Thus x, y, z are in G.P.

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18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…

Solution:

The given sequence is 8, 88, 888, 8888…

This sequence is not a G.P. However, it can be changed to G.P. as

Sn = 8 + 88 + 888 + 8888 + …………….. to n terms

= ⁸/₉ [9 +99 + 999+ 9999+ … to n terms]

= ⁸/₉ [(10-1) +(100-1) + (1000-1) + (1000-1) + … to n terms]

= ⁸/₉ [(10 +10² + 10³ + … to n terms)-(1+1+1+…n terms)]

= ⁸/₉ [^{10(10^n-1)}/₉ – n]

= ⁸⁰/₈₁ (10ⁿ – 1) – ⁸/₉ n

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19. Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, ¹⁄₂ .

Solution:

Required sum = 2×128 + 4×32 + 8×8 +16×2 +32x¹⁄₂

=64[4+2+1+¹/₂ +¹/₂²]

Here 4, 2, 1, ¹/₂, ¹/₂² is a G.P.

First term a = 4

Common ratio r = ¹/₂

Required sum = 64(³¹/₄) = 16(31) = 496

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20. Show that the products of the corresponding terms of the sequnnces form a, ar, ar²,…ar^n-1 and A, AR,AR²,…,AR^n-1 a G.P. and find the common ratio.

Solution:

It has to be proved that the sequence: aA, arAR, ar²AR²,…ar^n-1AR^n-1 forms a G.P.

^{Second term}/_{first term} = ^arAR/_aA = rR

^{Third term} /_secondterm = ar²AR²/_arAR = rR

Thus the above sequence forms a G.P. and the common ratio is rR

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For continued problems click here

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